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Oduvanchick [21]
2 years ago
8

Which is not an example of a scalar?

Physics
1 answer:
Crank2 years ago
3 0

Answer:

b) 2ft/s

Explanation:

A scalar has only magintude, not direction

6.2m, 3kg, and -100 o C are all scalars because they only have magnitude.

2ft/s is not a scalar because it has a direction.

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Planet X is in a stable circular orbit around a star, as shown in the figure. Which of the following graphs best predicts the an
damaskus [11]

Answer:

C

Explanation:

Angular momentum is the product of moment of inertia and angular velocity.

L = I × ω

Since the planet follows a stable circular orbit, I and ω are constant and non-zero.  Therefore, the angular momentum is constant and non-zero.

5 0
3 years ago
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8 0
3 years ago
The roller-coaster car shown in fig. 6-41 (h1 = 45 m, h2 = 16 m, h3 = 26 m), is dragged up to point 1 where it is released from
kirill [66]

There are many ways to solve this but I prefer to use the energy method. Calculate the potential energy using the point then from Potential Energy convert to Kinetic Energy at each points.

PE = KE

From the given points (h1 = 45, h2 = 16, h<span>3  </span>= 26)

Let’s use the formula: 

v2= sqrt[2*Gravity*h1]  where the gravity is equal to 9.81m/s2

v3= sqrt[2*Gravity*(h1 - h3 )] where the gravity is equal to 9.81m/s2

v4= sqrt[2*Gravity*(h1 – h2)] where the gravity is equal to 9.81m/s2

Solve for v2

v2= sqrt[2*Gravity*h1]      

    = √2*9.81m/s2*45m

v2= 29.71m/s

v3= sqrt[2*Gravity*(h1 - h3 )   

    =√2*9.81m/s2*(45-26)

    =√2*9.81m/s2*19 

v3=19.31m/s

v4= sqrt[2*Gravity*(h1 – h2)]        

    =√2*9.81m/s2*(45-16)

    =√2*9.81m/s2*(29)

v4=23.85m/s

7 0
3 years ago
A rock thrown horizontally from a bridge. Show that the height of the bridge hits the water below. The rock travels in a smooth
drek231 [11]

Answer:

This situation is related to parabolic motion and the main equation is:

y=y_{o}+V_{oy} t-\frac{gt^{2}}{2}   (1)

Where:

y=0 is the final height of the rock, asuming the top of the bridge touches the surface of the water

y_{o}  is the initial height of the rock

V_{oy}=0 is the vertical component of the initial velocity (it is zero because the rock was thrown horizontally)

t is the time the parabolic motion lasts

g  is the acceleration due gravity

Rewritting (1) with these conditions:

0=y_{o}+(0) t-\frac{gt^{2}}{2}   (2)

Hence:

y_{o}=\frac{gt^{2}}{2}  

8 0
4 years ago
Pls help i’ll give brainliest if you give a correct answer!!
Sonja [21]

Answer:

applied force since she is pushing the box

maybe, I'm not 100% sure

7 0
3 years ago
Read 2 more answers
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